TrueTears question thread

[kamil9876]

because the run is the x value at Q minus the x value at S. It makes a lot of sense when u draw the diagram of the three points and the lines

[TrueTears]

but how do you know the x value of Q is x?

[kamil9876]

i just called the point Q (x,y). Shouldve maybe added a subscript Q for it to make sense hah

[/0]

I used vectars:

Let

[kamil9876]

hah yep that's what i did but without the notation. Linear equation, vectors and similair triangles are all similair in some cases (no pun intended)

[TrueTears]

thanks i get i XD

yeah i did it another way and still got it lol

[TrueTears]

another question.

1. a) Sketch the hyperbola clearly indicating its centre vertices and asymptotes.

This i can do.

b) By setting up a suitable quadratic equation, show that the line is a tangent to the hyperbola .

This i can do.

c) Find the coordinates of the point of intersection of the tangent and the hyperbola . Let this point be A.

This i can do.

d) The line passes through the point B(3,2). By referring to the sketch you drew in part a, state the equation of the other tangent to the hyperbola that passes through B, and state the coordinates of the point of intersection, C, of this tangent with the hyperbola.

lol stuck on part d)

[/0]

I 'think' the other tangent is x = 3. Not sure though; although, it would fit the information.

[TrueTears]

yeah , but how would you solve it ?

[TrueTears]

Any ideas /0?

[TrueTears]

actually /0, you know your vectors way of doing that question?

I used vectars:

Let

i know it works but... the question says SQ = 3SP meaning the magnitudes , where as you had them as vectors? If the magnitude of SQ is 3 times SP , does that also mean the vector SQ is also 3 times vector SP?

[/0]

What do you mean by 'solve' it?


CAN SOMEONE ELSE PLEASE CONFIRM THIS

I just kinda looked at the graph and saw x=3 as a possible tangent through (3,2). Is that how you're supposed to do it? Maybe? Seeing as it says "refer to your graph in part a)".

So C = (3,0)

i know it works but... the question says SQ = 3SP meaning the magnitudes , where as you had them as vectors? If the magnitude of SQ is 3 times SP , does that also mean the vector SQ is also 3 times vector SP?

"The line segment SP is extended to the point Q where P is between S and Q and SQ = 3SP."

Since the line is extended, S, Q and P must be collinear, so you can use vectors.

[TrueTears]

What do you mean by 'solve' it?


CAN SOMEONE ELSE PLEASE CONFIRM THIS

I just kinda looked at the graph and saw x=3 as a possible tangent through (3,2). Is that how you're supposed to do it? Maybe? Seeing as it says "refer to your graph in part a)".

So C = (3,0)

i know it works but... the question says SQ = 3SP meaning the magnitudes , where as you had them as vectors? If the magnitude of SQ is 3 times SP , does that also mean the vector SQ is also 3 times vector SP?

"The line segment SP is extended to the point Q where P is between S and Q and SQ = 3SP."

Since the line is extended, S, Q and P must be collinear, so you can use vectors.

nice i see, but even if they are collinear if their magnitudes have that relation ship doesn't mean the vectors also share that relationship? (SQ = 3SP)

[TrueTears]

and also for that tangent question, yeah... i mean is there a more systematic way of doing it? Instead of just kinda guessing and looking at graph?

[kamil9876]

What do you mean by 'solve' it?


CAN SOMEONE ELSE PLEASE CONFIRM THIS

I just kinda looked at the graph and saw x=3 as a possible tangent through (3,2). Is that how you're supposed to do it? Maybe? Seeing as it says "refer to your graph in part a)".

So C = (3,0)

i know it works but... the question says SQ = 3SP meaning the magnitudes , where as you had them as vectors? If the magnitude of SQ is 3 times SP , does that also mean the vector SQ is also 3 times vector SP?

"The line segment SP is extended to the point Q where P is between S and Q and SQ = 3SP."

Since the line is extended, S, Q and P must be collinear, so you can use vectors.

nice i see, but even if they are collinear if their magnitudes have that relation ship doesn't mean the vectors also share that relationship? (SQ = 3SP)

But like /0 said, the line is extended by a factor of 3, meaning that the gradient remains unchanged. ANother way of looking at it is imagine the line y=3x, now there exists a segment of this line that has magnitude one (an example is this line with a domain of . This line segment can also be described by the vector i + j. While this line extended by a factor of 3 is like the function y=3x for domain which is the same as the vector i + j.

Moral of the story: Vectors are like line segments, where the i component is the run, while the j is the rise. A vector being extended by a factor of 3 is like a line segment being extended, the rise over run doesn't change and so the factor that the rise was extended is the same as the factor that the run was extended. This factor is equal to the factor that the magnitude was extended, this can be proven using pythatgoras theorem or even more elegantly by imagining 3 coppies of the vector being placed head to tail. e.g: draw some vector a now the vector 3a=a + a + a so it's like 3 coppies being attatched head to tail. Now the new hypotenuse is the sum of the three individual hypotenuses, while the run is the sum of the three runs, and the rise is the sum of the three rises. This is explained in teh attatched image.